Reverse Integer

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本文介绍了一种用于反转整数的有效算法，并提供了两种实现方法。讨论了处理边界情况的方法，例如输入数字为0或反转后可能导致整数溢出的情况。

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

Have you thought about this?

Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

很简单的一道题。

如下：

public class Solution {
    public int reverse(int x) {
      if(x == 0){
          return 0;
      }
      int fuhao = x >> 31 & 1;//1为负数
      if(fuhao == 1){
          x = -x;//转换成正数 
      }
      StringBuffer sb = new StringBuffer();
      while(x > 0){
          sb.append(x%10);
          x /= 10;
      }
      try{
        x = Integer.parseInt(sb.toString());
      }catch(Exception ex){
          return 0;//转换之后的数据溢出
      }
      if(fuhao == 1){
          x = -x;
      }
      return x;
    }
}

Runtime: 256 ms

注意题后的特殊情况，一个正常的数转换后可能出错，如input为1534236469，则retrun 0；

参考别人的代码，发现自己的思维固定在了Sstring上了，改进如下：

public class Solution {
    public int reverse(int x) {
      if(x == 0){
          return 0;
      }
      int fuhao = x >> 31 & 1;//1为负数
      if(fuhao == 1){
          x = -x;//转换成正数 
      }
      long x1 = 0;
      while(x > 0){
          x1 = x1 * 10 + x%10;
          x /= 10;
      }
      if(x1 > Integer.MAX_VALUE || x1 < Integer.MIN_VALUE){//转换失败
          return 0;
      }
      if(fuhao == 1){
          x = -(int)x1;
      }else{
          x = (int)x1;
      }
      return x;
    }
}

Runtime: 217 ms